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# 超急~四題物理題目求解釋!!

我只需要解釋，答案已給(除了第四題沒答案，不好意思)，盡量可以清楚一點，相當急，麻煩會這些題目的大大囉!! 謝謝^^1.A disk 8.00 cm in radius rotates at a constant rate of 1200 rev/min about its central axis. Determine:(a)its angular speed in radians per second.(b)the tangential speed at a point 3.00 cm from its center.(c)the radial acceleration of a point on the rim.(d)the total distance a point on the rim moves in 2.00 s.Answer: (a) 126 rad/s (b) 3.77 m /s (c) 1.26 km /s平方 (d) 20.1 m 2.A model airplane with mass 0.750 kg is tethered to the ground by a wire so that it flies in a horizontal circle 30.0 m in radius. The airplane engine provides a net thrust of 0.800 N perpendicular to the tethering wire.(a)Find the torque the net thrust produces about the center of the circle.(b)Find the angular acceleration of the airplane.(c)Find the translational acceleration of the airplane tangent to its flight path.Answer: (a) 24.0 N*m (b) 0.0356 rad/s平方 (c) 1.07 m /s平方3.The combination of an applied force and a friction force produces a constant total toque of 36.0 N*m on a wheel rotating about a fixed axis. The applied force acts for 6.00 s. During this time, the angular speed of the wheel increases from 0 to 10.0 rad/s.The applied force is then removed, and the wheel comes to rest in 60.0 s. Find:(a) the moment of inertia of the wheel(b) the magnitude of the torque due to friction(c) the total number of revolutions of the wheel during the entire interval of 66.0 s.Answer: (a) 21.6 kg *m平方 (b) 3.60 N*m (c) 52.5 rev4.A horizontal 800-N merry-go-round is a solid disk of radius 1.50 m and is started from rest by a constant horizontal force of 50.0 N applied tangentially to the edge of the disk. Find the kinetic energy of the disk after 3.00 s.

### 2 Answers

- 翻雷滾天 風卷殘雲Lv 79 years agoFavorite Answer
1a) 1200 rev is equal to 1200 x 2π = 2400π rad.

Hence the angular speed is ω = 2400π/60 = 40π = 126 rad/s

b) Tangential speed = rω = 0.03 x 40π = 3.77 m/s

c) Centripetal acc. = rω2 = 0.08 x (40π)2 = 1.26 km/s2

d) Tangential speed on the rim = 0.08 x 40π = 10.05 m/s

Hence total distance moved in 2 s = 10.05 x 2 = 20.1 m

2a) 0.8 x 30 = 24 Nm

b) Linear acc. = 0.8/0.75 = 1.07 m/s2

So angular acc. = 1.07/30 = 0.0356 rad/s2

c) 0.0356 x 30 = 1.07 m/s2

3a) Angular acc. of the wheel = 10/6 = 5/3 rad/s2

So moment of inertia = 36/(5/3) = 21.6 kg m2

b) Angular decelaration of the wheel = 10/60 = 1/6 rad/s2

Hence torque dur to friction = 21.6 x 1/6 = 3.6 N m

c) Angular displacement in the period of acc. is:

(1/2) x (5/3) x 102 = 250/3 rad

Angular displacement in the period of dec. is:

10 x 60 - (1/2) x (1/6) x 602 = 300 rad

So total angular displacement = 300 + 250/3 = 1150/3 rad

With no. of rev = 1150/(3 x 2π) = 52.5

4) Torque produced = 50 x 1.5 = 75 N m

Mass of the round = 80 kg, with moment of inertia = 0.5 x 80 x 1.52 = 90 kg m2.

So angular acc. = 75/90 = 5/6 rad/s2

After 3 s, angular velocity = 5/6 x 3 = 2.5 rad/s

So k.e. = 0.5 x 90 x 2.52 = 281.25 J

Source(s): 原創答案